Question

A circular conducting loop of radius $r_0$ and having resistance per unit length $\lambda$ is placed in a magnetic field $B$ which is constant in space and time as shown in the figure. The ends of the loop are crossed and pulled in opposite directions with a velocity $v$ such that the loop always remains circular and the radius of the loop goes on decreasing, then :

• A. Radius of the loop changes with time $t$ as $r=r_0-\frac{vt}{\pi }$
• B. $\text{EMF}$ induced in the loop as a function of time is $E=2Bv[r_0-\frac{vt}{\pi }]$
• C. Current induced in the loop is $I=\frac{Bv}{2\pi \lambda }$
• D. Current induced in the loop is $I=\frac{Bv}{\pi \lambda }$

Answer 1

Answer: (A), (B), (D)

Current induced in the loop is $I=\frac{Bv}{\pi \lambda }$

As, given radius of the loop is decreasing continuously, as a result perimeter of the loop is decreasing at a rate of $2v$

$\therefore \frac{d}{dt}\left(2\pi r\right)=2v$

$\frac{dr}{dt}=\frac{v}{\pi }$

$\Rightarrow dr=\frac{v}{\pi }dt$

Integrating both side with proper limits

${\int }_{r_0}^{r}dr={\int }_0^t\frac{v}{\pi }dt$

$\therefore r=r_0-\frac{v}{\pi }t$

Magnetic flux passes through the coil is given by

$\varphi =BA\cos 0^{\circ }=B\left(\pi r^2\right)$

Applying Faraday's law for induced emf

$E=\left|\frac{-d\varphi }{dt}\right|=B\cdot 2\pi r\frac{dr}{dt}$

$E=2B\pi (r_0-\frac{v}{\pi }t)\frac{v}{\pi }(\because dr/dt=v/\pi )$

$\therefore E=2Bv(r_0-\frac{v}{\pi }t)$

So, induced current in the loop will be

$I=\frac{E}{R}=\frac{2Bv(r_0-\frac{v}{\pi }t)}{\lambda [2\pi (r_0-\frac{v}{\pi }t)]}=\frac{Bv}{\pi \lambda }$

Therefore, options $\left(A\right),\left(B\right)$ and $\left(D\right)$ are the correct choice.